550 PROMISCUOUS EXERCISES. And when x = 3 Ac, then z 886 of a foot, or 103 Again, to find the time, after the above, when EF arrives at cp, or when the water in the ditches G The notation remaining as before, then 2bz g(x-2) per sec. runs through AF, B and gb(3-2)g(x-2) per sec. thro' ED nearly; A conseq. 2b √☎ × (12 + z) √ (x − z) = v is the velocity per 5A Assume z=Ax2+вx2+cx2+DÃ2 &c. So shall A A2+4B 3 ▲3+4AB+8c 23 √ (x − x) = x 12+z=12+ax2+вx2+cx2 &c.; x2 &c.; 16 (12+z). √/(x−z). x=12x3 x − 6ax3x— (§ A2+6в)x3⁄4x &c. ; 1 3 6A3 Then, equating the like terms, &c. we have But by the first process, when x=3, z=•886; which substitated for them, we have z = •886, and the series 1.63: therefore the correct fluents are And when z=3= AC, it gives x = 6·369 for the height of the tide without, when the ditches are filled to the top of the sluice, or 3 feet high; which answers to 3h 11′ 4′′. . Lastly, . Lastly, to find the time of rising the remaining 3 feet above the top of the sluice; let *=ce the height of the tide above co, z=CE ditto in the ditches above CD ; and the other dimensions as before. G GI C A Then/g: /EG: : 2g : 2 √✓ g (x− z) = the A of the water in the ditches, per second; hence v:ż:: 2 3, x = 7; the heights above the top of the sluice; answering to 6 and 10 feet above the bottom of the ditches. That is, for the water to rise to the height of 6 feet within the ditches, it is necessary for the tide to rise to 10 feet without, which just answers to 5 hours; and so long it would take to fill the ditches 6 feet deep with water, their horizontal area being 200000 square feet. + Further, when x = 6, then z = 2.117 the height above *Note. The fluxional equation mi — ż√(x—2) may be integrated without series,-EDITOR, the top of the sluice; to which add 3, the height of the Note. In the foregoing problems, concerning the afflux of Also in some places, where the difference between two pa- We may further add a remark here concerning the method z = x2 + x2+s+x+2s &c. for ascending, terms terms equal, which will give the value of the common differ- ་ Thus, for the last of the three fluxional equations above s—1X 2 x Again, for the 2d equation mzor z = (12 + z) √ (x − z) ∞ Lastly, assuming the same form of series for z and z as in : / X xr 1 which mult. by 2x+z, gives 2x3x--xr+2x &c. : here equat- Such then is the regular and legitimate way of proceeding, VOL. II 71 PROBLEM : PROBLEM 53. To determine the Fall of the Water in the Arches of a Bridge. The effects of obstacles placed in a current of water, such as the piers of a bridge, are, a sudden steep descent, and an increase of velocity in the stream of water, just under the arches, more or less in proportion to the quantity of the obstruction and velocity of the current; being very small and hardly perceptible where the arches are large and the piers few or small, but in a high and extraordinary degree at London-bridge, and some others, where the piers and the sterlings are so very large, in proportion to the arches. This is the case, not only in such streams as run always the same way, but in tide rivers also, both upward and downward, but much less in the former than in the latter. During the time of flood, when the tide is flowing upward, the rise of the water is against the under side of the piers; but the difference between the two sides gradually diminishes as the tide flows less rapidly towards the conclusion of the flood. When this has attained its full height, and there is no longer any current, but a sțillness prevails in the water for a short time, the surface assumes an equal level, both above and below bridge. But as soon as the tide begins to ebb or return again, the resistance of the piers against the stream, and the contraction of the waterway, cause a rise of the surface above and under the arches, with a full and a more rapid descent in the contracted stream just below. The quantity of this rise, and of the consequent velocity below, keep both gradually increasing, as the tide continues ebbing, till at quite low water, when the stream or natural current being the quickest, the fall under the arches is the greatest. And it is the quantity of this fall which it is the object of this problem to determine. 1 Now, the motion of free running water is the consequence of and produced by the force of gravity, as well as that of any other falling body. Hence the height due to the velocity, that is, the height to be freely fallen by any body to acquire the observed velocity of the natural stream, in the river a little way above bridge, becomes known. From the same velocity also will be found that of the increased current in the narrowed way of the arches, by taking it in the reciprocal proportion of the breadth of the river above, to the contracted way in the arches; viz. by saying, as the latter is to the former, so is the first velocity, or slower motion, to the quicker. Next, from this last velocity, will be found the height due to it |